Problem: Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$. Find the value of
\[\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}.\]
Explanation: We will examine the first term in the expression we want to evaluate, $\frac{\sin 2x}{\sin 2y}$, separately from the second term, $\frac{\cos 2x}{\cos 2y}$. Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$, we have $$\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}.$$Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2. To use the identity $\sin^2\theta + \cos^2\theta = 1$, we will cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$. Equation 1 then becomes $\sin x = 3\sin y$. Equation 2 then becomes $\cos x = \frac{1}{2} \cos y.$ We can square both of the resulting equations and match up the resulting LHS with the resulting RHS and get $$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y.$$Applying the identity $\cos^2 y = 1 - \sin^2 y$, we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into $$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y.$$Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y $. Therefore, $\sin^2 y = \frac{3}{35}$. Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$), we can solve for $\sin^2 x$ as follows: $$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}.$$Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$, we can solve for $\frac{\cos 2x}{\cos 2y}$:
\begin{align*}
\cos 2x &= 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}, \\
\cos 2y &= 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}.
\end{align*}Hence, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$.

Finally,
\[\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \boxed{\frac{49}{58}}.\]